Abstract. An M/M/1 queue with delayed vacation is studied. ..... Letting M denote the number of B1 + D events before the server takes a vacation, and using the ...
M/M/1 Queueing System with Delayed Controlled Vacation Yonglu Deng∗, Zhongshan University W. John Braun† , University of Winnipeg Yiqiang Q. Zhao†, University of Winnipeg
Abstract An M/M/1 queue with delayed vacation is studied. If the server has been idle for a period of time (called the delay time), the server begins an exponentially distributed vacation which is repeated as long as the number of customers in the system remains less than some number K. For the cases of exponential and deterministic delay time, exact expressions for the steady state probability distribution are obtained, together with associated performance measures. System optimization is also considered; values of K are given which minimize the average total cost per unit time, and it is shown that the optimal delay period is either 0 (no delay) or infinite (no vacation), in case of Poisson arrivals. Key words: Queueing model, stationary distribution, delayed vacation.
∗
Department of Mathematics, Zhongshan University, Guangzhou, 510275, China Department of Mathematics and Statistics, University of Winnipeg, Winnipeg, Manitoba, Canada, R3B 2E9. †
1
1
Introduction
Queueing systems with server’s vacations have been studied extensively. A comprehensive review on vacation models, methods and results up to 1991 can be found in either Doshi [6] or Takagi [11]. In the last few years, increasing interest in studying queueing systems with various rules of vacation has led to many extensions of previously existing results. For example, a batch arrival model with a finite capacity for the buffer size can be used to model some telecommunications systems using a time division multiple access (TDMA) scheme (Frey and Takahashi, [7]). Researchers have also done performance analysis on systems where probability distributions of the variables are more general and closer to reality (e.g. Chao and Zhao [2]). In this paper, we consider vacation models which are extended in two different ways. The first one is the queueing system in which the length of the vacations can be controlled by means of the number of customers K arriving to the system during the vacation, and the level K may be chosen according to the arrival rate and the service rate, the cost per unit of waiting time and the cost of the server being transferred from vacation to work. Secondly, we allow for a delay time before a vacation begins. During the delay time, the server is situated in warm standby state and it will start service immediately upon arrival of a customer to the system. It seems that the consideration of such a delayed vacation might be reasonable if the cost of a server’s warm switch-on is lower than that of a server’s cold switch-on. The case of delayed vacation has also very recently been studied by Frey and Takahashi [8] and Sakai et al [9] where the term close-down time is used. They consider general service, delay and vacation times with Poisson (batch) arrivals, but must rely on numerical examples to gain insights into their results. We focus on a simpler model with exponential service and vacation which allows us to gain greater insight into the system with less effort. In particular, we are able to consider optimization of the system with respect to the control parameter and the delay parameter. We show that for Poisson arrivals and exponential service and vacation, there is no nontrivial optimal delay period. However, the use of the control parameter to extend the vacation period until K customers arrive in the system can result in improved performance of the system.
2
2
Description and Analysis of the Model
We make the following assumptions. 1. Customers arrive at the system according to a Poisson process with intensity λ and there is one server in the system. The queue discipline is FCFS. 2. Service times are assumed to be exponentially distributed with mean 1/µ. The traffic intensity is ρ = λ/µ < 1. 3. A period of time called the delay time occurs before the server goes on vacation. The delay can be interrupted by the arrival of a customer in which case the server resumes service. Thus, the length of the delay time is min(X, Y ) where X is exponentially distributed with mean 1/λ, and Y is a random variable with mean 1/θ1 . In this paper, we assume that Y is either exponentially distributed or deterministic. 4. If the delay time is completed before the arrival of a customer, the server begins a vacation whose length is exponentially distributed with mean 1/θ2 . Upon completion of a vacation, the server resumes service if K ( ≥ 1) or more customers are in the system. Otherwise, it takes another vacation having length independent of and identically distributed to the preceding one. 5. All aforementioned random variables are independent of each other. Let S(t) = 0, S(t) = 1, and S(t) = 2 denote the events that the server is busy, in delay period, and on vacation at epoch t, respectively. Define p0j (t) = P (N (t) = j, S(t) = 0)
(j = 1, 2, 3, . . .)
p1 (t) = P (N (t) = 0, S(t) = 1) p2j (t) = P (N (t) = j, S(t) = 2)
(j = 0, 1, 2, . . .)
N (t) will denote the number of customers in the system at time t.
3
2.1
Exponential Delay Case
When the delay time is exponentially distributed, {(N (t), S(t)) , t ≥ 0} is a standard continuoustime Markov chain. From the theory of Markov chains (e.g. [3]), it follows that {(N (t), S(t)) , t ≥ 0} has a unique equilibrium distribution which satisfies the following family of equations. Setting p0j = lim p0j (t)
(j = 1, 2, 3, . . .)
t→∞
p1 = lim p1 (t) t→∞
p2j = lim p2j (t)
(j = 0, 1, 2, . . .)
t→∞
we have λp20 = θ1 p1 λp2j = λp2,j−1
(j = 1, 2, 3, . . . , K − 1)
(λ + θ2 )p2j = λp2,j−1
(j = K, K + 1, K + 2, . . .)
(λ + θ1 )p1 = µp01
(1)
(λ + µ)p01 = λp10 + µp02 (λ + µ)p0j = λp0,j−1 + µp0,j+1
(j = 2, 3, 4, . . . , K − 1)
(λ + µ)p0j = λp0,j−1 + µp0,j+1 + θ2 p2j ∞ X
p0j + p1 +
j=1
∞ X
(j = K, K + 1, K + 2, . . .)
p2j = 1
j=0
The solution of these equations is as follows (see Appendix for a derivation):
p1 =
p2j =
θ1
λ
λθ2 (1 − ρ) λθ1 + λθ2 + Kθ1 θ2
p1
λ λ+θ2
(j = 0, 1, 2, . . . , K − 1) j−k+1
p0j = p1 ρj +
p20 (j = K, K + 1, K + 2, . . .)
j−1 X
ρj−i p2i
i=0
4
(j = 1, 2, 3, . . .)
(2)
(3)
(4)
2.2
Deterministic Delay Case
When the delay time is deterministic, {(N (t), S(t)) , t ≥ 0} is not a Markov chain, but it can be extended to a continuous-state Markov chain with the use of the supplementary variable technique (e.g. [5], [4]). In particular, we introduce the random variable X(t) which denotes the elapsed delay time at time t. Setting p0j = lim p0j (t)
(j = 1, 2, 3, . . .)
t→∞
p1 (x)dx = lim P (x ≤ X(t) < x + dx, S(t) = 1), t→∞
p2j = lim p2j (t)
0 < x < 1/θ1
(j = 0, 1, 2, . . .)
t→∞
we have λp20 = p1 (1/θ1 ) λp2j = λp2,j−1
(j = 1, 2, 3, . . . , K − 1)
(λ + θ2 )p2j = λp2,j−1
(j = K, K + 1, K + 2, . . .)
p1 (0) = µp01 (λ + µ)p01 = λ
Z
(5)
1/θ1
0
p1 (x)dx + µp02
(λ + µ)p0j = λp0,j−1 + µp0,j+1
(j = 2, 3, 4, . . . , K − 1)
(λ + µ)p0j = λp0,j−1 + µp0,j+1 + θ2 p2j d p1 (x) = −λp1 (x), 0 < x < 1/θ1 dx ∞ X
p0j +
j=1
Z
1/θ1
0
p1 (x)dx +
(j = K, K + 1, K + 2, . . .)
∞ X
p2j = 1
j=0
The solution in this case is as follows: (eλ/θ1 − 1)(1 − ρ)θ2 p1 = λ − θ2 − eλ/θ1 θ2 − Kθ2
p2j =
(1−ρ)θ2 λ/θ
1 θ2 +Kθ2 λ−θ2 +ej−k+1
λ λ+θ2
p20
j
p0j = p1 ρ +
j−1 X
(j = 0, 1, 2, . . . , K − 1) (j = K, K + 1, K + 2, . . .)
ρj−i p2i
i=0
5
(j = 1, 2, 3, . . .)
2.3
Mean Queue Length
We are now in a position to deduce the mean numbers of customers in the system and in the queue. Let L denote the mean number of customers in the system, and let Lq denote the mean queue length. Then, we have
and
!2
Kλ + θ2
!2
Kλ + θ2
ρ λ θ1 θ2 K(K − 1) L= + + 1 − ρ λθ1 + λθ2 + Kθ1 θ2 2 θ2
K(K − 1) λ θ1 θ2 ρ2 + + Lq = 1 − ρ λθ1 + λθ2 + Kθ1 θ2 2 θ2
(6)
(7)
in the exponential delay case. In the deterministic delay case, we have L=
and Lq =
2.4
ρ 2 λ 2 + 2 K λ θ 2 − K θ2 2 + K 2 θ2 2 + λ 1−ρ 2 θ 2 λ − θ 2 + e θ1 θ 2 + K θ 2 2 λ 2 + 2 K λ θ 2 − K θ2 2 + K 2 θ2 2 ρ2 + λ 1−ρ θ1 2 θ2 λ − θ2 + e θ2 + K θ2
(8)
(9)
Waiting Time
Let Wq denote the virtual waiting time of customers arriving in the system. Then p20 λµ K(K − 1)µp20 λ(p1 + p20 K) λ2 E[Wq ] = −W (0) = + Kµ + + (10) + (µ − λ)θ2 θ2 µ−λ 2λ(µ − λ) (µ − λ)2 ∗0
#
"
and 1 E[Wq |Wq > 0] = 1 − p1
λ2 K(K − 1)µp20 λ(p1 + p20 K) p20 λµ + + Kµ + + (µ − λ)θ2 θ2 µ−λ 2λ(µ − λ) (µ − λ)2 (11) "
#
!
We also note that Little’s formula also holds for this model.
2.5
Mean Length of Operational Period
The period of time from the beginning of a delay to the end of the subsequent delay will be defined as an operational period for the system; we denote its length by A. 6
By the model assumptions, a busy period which begins with one customer in the system, or a vacation of length VK with a control level K will begin as soon as the delay terminates depending on whether a customer arrives or the server goes on vacation. In the first case, the busy period of length B1 will be followed (eventually) by another delay of length D and then the above process will be repeated. In the latter case, a busy period of length BR will follow the vacation of length VK where R is a random variable with values in {K, K + 1, . . .}. Another delay of length D follows BR so that the above process will then be repeated. From the above considerations, it is clear that an operational period consists of a delay D, a random number (say, M , a non-negative integer-valued random variable) of successive B1 + D events, a vacation of length VK , and a busy period of length BR . In particular, we have E[A] = E[D] + E[M ](E[B1 ] + E[D]) + E[VK ] + E[BR ]
(12)
The mean length of a delay is easily found. That is, D = min(X, Y ) where X is an exponential random variable with mean 1/λ, and Y is a random variable with mean 1/θ1 . If Y is exponential, then D is an exponential random variable with mean E[D] = 1/(λ + θ1 )
(13)
If Y is deterministic, then D is a truncated exponential random variable with mean E[D] = (1 − e−λ/θ1 )/λ
(14)
Letting M denote the number of B1 + D events before the server takes a vacation, and using the memoryless property of the exponential distribution, it is seen that these events are independent, and they end with a vacation with a certain probability p0 . Therefore, P (M = m) = p0 (1 − p0 )m so that E[M ] =
7
(m = 0, 1, 2, . . .)
1 − p0 p0
(15)
When the delay time Y is exponential, p0 = P (X > Y ) =
θ1 λ + θ1
and when Y is deterministic, p0 = P (X > Y ) = e−λ/θ1 It is also clear that a busy period beginning with r customer in the system has mean E[Br ] =
r µ−λ
(r = 1, 2, 3, . . .)
(16)
It is shown in the appendix that the mean busy period beginning with a random number (R) of customers is given by E[BR ] =
λ + Kθ2 (µ − λ)θ2
(17)
A vacation period with control parameter K has mean length given by E[VK ] =
1 K + λ θ2
(18)
Substituting (13), (15), (18), and (17) into (12) gives, upon simplification, E[A] =
λ(θ1 + θ2 ) + Kθ1 θ2 (1 − ρ)λθ1 θ2
(19)
in the exponential case, and using (14) in place of (13) gives E[A] =
θ2 eλ/θ1 + (K − 1)θ2 + λ λ(1 − ρ)θ2
in the deterministic case.
3
Optimization of the Control Parameter
Suppose that the cost per unit time is • C0 ≤ 0 for server vacation • C1 ≥ 0 for delay (warm standby) 8
(20)
• C2 ≥ 0 for normal service
(C2 ≥ C1 generally)
• C4 ≥ 0 per customer waiting in the system In addition, C3 ≥ 0 denotes the cost due to switching the server from vacation to normal service. For an operational period, the expected cost due to vacation is given by C0 (K/λ + 1/θ2 ) The expected cost due to delay is C1 E[D]/p0 The expected cost while the server is busy is C2
λ + θ2 K (µ − λ)θ2
!
C2 + µ−λ
!
1 −1 p0
From the previous section, the expected length of an operational period is K 1 E[D] λ + θ2 K 1 E[A] = + + + + λ θ2 p0 (µ − λ)θ2 µ − λ
!
1 −1 p0
In the case of exponential delay, we then find that the cost for an operational period is given by 1 ρλ(θ1 + θ2 ) + Kθ1 θ2 K C1 + C = C0 + C2 + C3 + λ θ2 θ1 (1 − ρ)λθ1 θ2 so that the average cost per unit time is
C0 (1 − ρ)θ1 (Kθ2 + λ) + C1 (1 − ρ)λθ2 + C2 (ρλ(θ1 + θ2 ) + Kρθ1 θ2 ) + C3 (1 − ρ)λθ1 θ2 C = E[A] λ(θ1 + θ2 ) + Kθ1 θ2 The average total cost per unit time is then G=
C + C4 L E[A]
Setting dG/dK = 0, we obtain θ1 + θ2 C4 θ1 θ2 K 2 +C4 λ(θ1 +θ2 )K+C0 (1−ρ)λθ2 −C1 (1−ρ)λθ2 −C3 (1−ρ)λθ1 θ2 +C4 λ λ − 2 2 The nonnegative solution of this is given by
K ∗ = (θ1 θ2 )−1 −λ(θ1 + θ2 ) + 9
s
θ1 θ2 T λ2 (θ1 + θ2 )2 − 2 C4
!
=0
where
θ1 + θ2 T = C0 (1 − ρ)θ2 − C1 (1 − ρ)θ2 − C3 (1 − ρ)θ1 θ2 + C4 λ λ − 2
!
K ∗ is nonnegative if T ≤ 0, and the second derivative d2 G/dK 2 is nonnegative in this case. Therefore, if Kmin is the nearest integer to K ∗ , then Kmin will minimize cost (assuming all other quantities are held constant). A similar analysis applies In the case of deterministic delay; again, optimization with respect to K gives rise to a quadratic equation in K for which the nonnegative solution is given by ∗
K =1−e
λ θ1
λ − + θ2
q
2λ
C4 (λ2 µ2 + λ µ2 θ2 ) + 2T + C4 e θ1 µ2 θ2 2 + 2 C3 λ µ2 θ2 2 q
C4 µ θ 2
where λ
λ
λ
T = λ 2 θ 2 2 − e θ1 λ 2 θ 2 2 − λ µ θ 2 2 − C0 λ µ θ 2 2 + e θ1 λ µ θ 2 2 + C0 e θ1 λ µ θ 2 2 − C3 λ 2 µ θ 2 2 λ
λ
+C0 µ2 θ2 2 − C0 e θ1 µ2 θ2 2 − C4 e theta1 µ2 θ2 2 Remark:
A similar treatment will yield optimality conditions on θ1 and θ2 . We note
that the cost function is monotonic in θ1 , so that the optimal delay period for this system is either infinite (no vacation) or 0 (no delay). We conjecture that this behaviour is a result of the memoryless property of the arrival process. On the basis of a simulation study, we have observed that, for non-Poisson arrivals, a nontrivial delay period will be optimal.
Appendix A.1 Derivation of Stationary Queue Length Distribution To solve the family of equations (1), we introduce the following generating functions G0 (s) =
∞ X
p0j sj
(|s| ≤ 1)
p2j sj
(|s| ≤ 1)
j=1
and G2 (s) =
∞ X
j=0
10
From the first three equations of (1), it is clear that p20 = p21 = · · · = p2,K−1
and p2j =
λ λ + θ2
!j−k+1
(21)
p20 =
θ1 p1 λ
(22)
p20
(j = K, K + 1, K + 2, . . .)
(23)
from which it follows that we can write p20 θ2 (1 − sk ) G2 (s) = λ+ λ + θ2 − sλ 1−s
!
(24)
Multiplying the 5th, 6th and 7th equations of (1) by sj and summing over j ∈ {1, 2, 3, . . .} gives (λ + µ)
∞ X
p0j = s(λp1 + µp02 ) +
j
s (λp0,j−1 + µp0,j+1 ) +
j=2
j=1
or
∞ X
∞ X
sj θ2 p2j
j=K
(λ + µ)G0 (s) = (λs + µ/s)G0 (s) + sλp1 − µp01 + θ2
∞ X
sj p2j
j=K
This, together with (23) and the 4th and 1st equations of (1), gives θ2 p20 sK λ λ + θ2 − sλ θ2 p20 sK λ = (s − 1)λp1 − λp20 + λ + θ2 − sλ λp20 = (s − 1)λp1 − λ + θ2 − sλ − θ2 sK λ + θ2 − sλ !! λp20 1 − sK = (s − 1) λp1 + λ + θ2 λ + θ2 − sλ 1−s
(λ + µ − λs − µ/s)G0 (s) = sλp10 − (λ + θ1 )p1 +
Dividing through by (s − 1) and using (24), we can write G0 (s)(µ/s − λ) = λ (p1 + G2 (s)) so that
λs (p1 + G2 (s)) µ − sλ In terms of the traffic intensity, this can be written as G0 (s) =
G0 (s) =
ρs (p1 + G2 (s)) 1 − sρ 11
(25)
Expanding in geometric series gives ∞ X
G0 (s) =
j
(ρs) p1 + G2 (s)
j=1
∞ X
(sρ)j
j=1
and expanding G2 (s) further gives G0 (s) =
∞ X
(ρs)j p1 +
∞ ∞ X X
si+j ρj p2i
j=1 i=0
j=1
from which we can see that j
p0j = p1 ρ +
j−1 X
ρj−i p2i
(j = 1, 2, 3, . . .)
(26)
i=0
It remains for us only to determine p1 . Letting s → 1 in (24) and (25), we obtain θ2 G2 (1) = (λ + Kθ2 )p20 and (µ − λ)G0 (1) = λG2 (1) + λp1 By using the 1st and 8th equations of (1), it follows that p1 =
λθ2 (1 − ρ) λθ1 + λθ2 + Kθ1 θ2
(27)
Thus, the equilibrium distribution of {(N (t), S(t)), t ≥ 0} is given by (21), (22), (23), (26) and (27).
A.2 Queue Length Distribution in the Deterministic Case Defining G0 (s) and G2 (s) as for the exponential case, we can again obtain (25) and (24) where we now define p1 =
Z
1/θ1
0
p1 (x)dx
These relations determine G0 (s) and G2 (s) in terms of p20 and p1 . Thus, (21), (23) and (26) again hold, but with p20 =
(1 − ρ)θ2 λ − θ2 + eλ/θ1 θ2 + Kθ2
p1 =
(eλ/θ1 − 1)(1 − ρ)θ2 λ − θ2 − eλ/θ1 θ2 − Kθ2
and
12
A.3 Derivation of Mean Queue Length Let N and Q denote the numbers of customers in the system and in the queue, respectively, assuming the system is in equilibrium. Also, let GN (s) and GQ (s) be the probability generating functions of N and Q, respectively. Clearly, GN (s) = G0 (s) + G2 (s) + p1
(|s| ≤ 1)
(28)
It is also easy to see that GQ (s) = G2 (s) + p1 − p01 + G0 (s)/s Using (25) and the 4th equation of (1), we have GQ (s) = G2 (s) + p1 +
λ + θ1 λs (G2 (s) + p1 ) − p1 s(µ − λs) µ
which, upon rearrangement, yields µ − λ − θ1 µ + λ − λs ρ G2 (s) + p1 + GQ (s) = µ − λs 1 − ρs µ
!
(29)
The means L = E[N ] and Lq = E[Q] are obtained by evaluating the derivatives G0N (1) and G0Q (1). To do this, we note that (24) can be differentiated easily to give G02 (1)
= p20
K(K − 1) λ2 + Kλθ2 + 2 θ22
!
(30)
Using (25), we can readily obtain G00 (1) =
λµ λ (G2 (1) + p10 ) + G0 (1) 2 (µ − λ) µ−λ 2
(31)
Since L = G00 (1) + G02 (1), we can use (30) and (31) to obtain (λ + Kθ2 )p20 λµ p + L= 1 (µ − λ)2 θ2
!
µp20 + µ−λ
K(K − 1) λ2 + Kλθ2 + 2 θ22
!
From this, we can obtain the special cases of exponential and deterministic delay displayed in equations (6), (7), (8) and (9). 13
A.4 Waiting Time Analysis Let Wq denote the virtual waiting time of customers arriving in the system, assuming the system is in equilibrium, and let Wq (x) denote the corresponding distribution function. Then
Wq (x) = P (Wq ≤ x, S = 2, N < K) + P (Wq ≤ x, S = 2, N ≥ K) + P (Wq ≤ x, S = 0) + p1 (32) Denote the first 3 terms on the right hand side by W21 (x), W22 (x) and W01 (x), respectively. We will evaluate E[Wq ] using ∗0 ∗0 ∗0 E[Wq ] = −W ∗0 (0) = − (W21 (0) + W22 (0) + W10 (0))
where W ∗0 (0) = lim W ∗0 (s) s→0
and W (s) denotes the Laplace-Stieltjes (L-S) transform of W (x). ∗
If a customer arrives at the system during the server’s vacation and sees N = j (< K) customers in the system, then the waiting time in the queue will be the sum of K − (j + 1) interarrivals, a residual vacation and j service times. Because of independence and the memoryless property of the exponential distribution, we have ∗0 W21 (s) =
=
Z
∞
0 K−1 X
e−sx dW21 (x) p2j
j=0
= p20 = p20
λ λ+s
!K−(j+1)
θ2 θ2 + s
! K−1 X
θ2 θ2 + s
!
j=0
λ λ+s
!
µ µ+s
!j
!K−(j+1)
µ µ+s
!j
θ2 θ2 + s
(λ + s)(µ + s) λ s(λ − µ) λ+s
!K
µ − µ+s
!K
(33)
If a customer arrives in the system during vacation and sees N = j (≥ K) customers in the system, then the waiting time in the queue is the sum of a residual vacation and j service times. Thus, ∗0 W22 (s)
=
Z
0
∞
e−sx dW22 (x) 14
∞ X
=
θ2 θ2 + s
p2j
j=K
= p20 = p20
θ2 θ2 + s
!
θ2 θ2 + s
!
!
µ µ+s
∞ X
j=K
!j
λ λ + θ2
!(j−K+1)
!K
1 −
µ µ+s
λ+θ2 λ
µ µ+s
!j
(34)
µ µ+s
Similarly, if a customer arrives in the system during a busy period and encounters N = j(≥ 1) customers in the system, then the waiting time in the queue is the sum of a residual service time and j − 1 service times. It then follows that ∗ W01 (s)
=
Z
∞ −sx
0
e
dW01 (x) =
∞ X
µ µ+s
p0j
j=1
!j
Using (26) we see that ∗ W01 (s)
=
∞ X
j
{p1 ρ +
j=1
j−1 X
p2i ρ
j−i
i=0
µ } µ+s
!j
Using (22) and (23) and several algebraic manipulations gives K µ 1 − p1 λ µ+s ∗ + + W01 (s) = p20 µ µ + s − λ p20 1 − µ+s
µ µ+s
!K
Combining (33), (34) and (35), we obtain
θ2 θ2 + s
W ∗ (s) = p20 +
1 λ p1 + µ + s − λ p20
λ λ+s
K
λ λ+s
− −
K µ µ+s µ µ+s
K µ − µ+s µ 1 − µ+s
Hence, we have (10).
+
+
λ(µ + s) (λ + θ2 )(µ + s) − λµ
K µ µ+s µ λ+θ2 − µ+s λ
K µ µ+s µ λ+θ2 − λ µ+s
(35)
p1 + p20
(36)
The probability that a customer arriving in the system has to wait for service is given by 1 − p1
(37)
Hence, given Wq > 0, the conditional distribution function Wq (x) is given by P (Wq ≤ x|Wq > 0) =
(W21 (x) + W22 (x) + W01 (x) 1 − p1
Then, in view of (32) and (10), we obtain (11). 15
(38)
A.4 Analysis of the Operational Period We consider the problem of obtaining the equation (17) for E[BR ], the expected length of a busy period beginning with R customers in the system. First, note that
P (R = r) =
In view of (23), we have
∞ X
j=K
−1
p2j
p2r
θ2 P (R = r) = λ
(r = K, K + 1, K + 2, . . .)
λ λ + θ2
!r−K+1
(39)
Because of (16), we can condition on R, and write E[BR ] = =
∞ X
r=K ∞ X
E[BR |R = r]P (R = r) E[Br ]P (R = r)
r=K
=
∞ X
r=K
r θ2 µ−λ λ
λ λ + θ2
!r−K+1
from which (17) follows immediately. We turn next to the derivation of the equation (18) for E[VK ], the expected length of vacation. Let X, Z and S be random variables which are independent and exponentially distributed with mean 1/λ, 1/θ2 and 1/µ, respectively. We use a recurrence method as follows: E[VK ] = E[VK |Z < X]P (Z < X) + E[VK |X ≤ Z]P (X ≤ Z) 0 = E[Z + VK0 |Z < X]P (Z < X) + E[X + VK−1 |X ≤ Z]P (X ≤ Z)
= E[Z|Z < X]P (Z < X) + E[X|X ≤ Z]P (X ≤ Z) 0 +E[VK0 ]P (Z < X) + E[VK−1 ]P (X ≤ Z) 1 θ2 λ = + E[VK ] + E[VK−1 ] λ + θ2 λ + θ2 λ + θ2
(40)
0 where VK0 and VK−1 have the same distributions as VK and VK−1 , respectively, but they are
independent of X and Z. We can then write E[VK ] =
1 + E[VK−1 ] λ 16
(41)
To obtain E[V1 ], we note that E[V1 ] = E[V1 |Z < X]P (Z < X) + E[V1 |X ≤ Z]P (X ≤ Z) = E[Z + V10 |Z < X]P (Z < X) + E[Z|X ≤ Z]P (X ≤ Z) θ2 = E[Z] + E[V1 ] λ + θ2 1 1 = + λ θ2
(42)
Combining (41) and (42), we obtain (18).
Acknowledgement This work was supported by research grants from the Natural Sciences and Engineering Research Council of Canada (NSERC).
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